Question

If the foot of the perpendicular from the point $A (-1,4,3)$ on the plane $P : 2 x + my + nz =$ 4 , is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point A from the plane $P$, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :

• A. 1
• B. $\sqrt{26}$
• C. $2 \sqrt{2}$
• D. $\sqrt{14}$

Answer 1

Answer: (B)

Let $B$ be foot of $\perp$ coordinates of $B =\left(-2, \frac{7}{2}, \frac{3}{2}\right)$
Direction ratio of line $AB$ is $\langle 2,1,3\rangle$ so $m=1, n=3$
So equation of $AC$ is $\frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}=\lambda$
So point $C$ is $(3 \lambda-1,-\lambda+4,-4 \lambda+3)$. But
$C$ lies on the plane, so
$ 6 \lambda-2-\lambda+4-12 \lambda+9=4 $
$ \Rightarrow \lambda=1 \Rightarrow C(2,3,-1) $
$ \Rightarrow A C=\sqrt{26}$